tenth edition - suranaree university of...
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VECTOR MECHANICS FOR ENGINEERS:
DYNAMICS
Tenth Edition
Ferdinand P. Beer
E. Russell Johnston, Jr.
Phillip J. Cornwell
Lecture Notes:
Brian P. SelfCalifornia Polytechnic State University
CHAPTER
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14Systems of ParticlesSystems of Particles
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Vector Mechanics for Engineers: Dynamics
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Contents
14 - 2
Introduction
Application of Newton’s Laws:
Effective Forces
Linear and Angular Momentum
Motion of Mass Center of System
of Particles
Angular Momentum About Mass
Center
Conservation of Momentum
Sample Problem 14.2
Kinetic Energy
Work-Energy Principle.
Conservation of Energy
Principle of Impulse and Momentum
Sample Problem 14.4
Sample Problem 14.5
Variable Systems of Particles
Steady Stream of Particles
Steady Stream of Particles.
Applications
Streams Gaining or Losing Mass
Sample Problem 14.6
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14 - 3
Engineers often need to analyze the dynamics of systems of
particles – this is the basis for many fluid dynamics
applications, and will also help establish the principles used
in analyzing rigid bodies
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Introduction
14 - 4
• In the current chapter, you will study the motion of systems
of particles.
• The effective force of a particle is defined as the product of
it mass and acceleration. It will be shown that the system of
external forces acting on a system of particles is equipollent
with the system of effective forces of the system.
• The mass center of a system of particles will be defined
and its motion described.
• Application of the work-energy principle and the
impulse-momentum principle to a system of particles will
be described. Result obtained are also applicable to a
system of rigidly connected particles, i.e., a rigid body.
• Analysis methods will be presented for variable systems
of particles, i.e., systems in which the particles included
in the system change.
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Application of Newton’s Laws. Effective Forces
14 - 5
• Newton’s second law for each particle Pi
in a system of n particles,
force effective
forces internal force external
1
1
ii
iji
iii
n
jijiii
ii
n
jiji
am
fF
amrfrFr
amfF
• The system of external and internal forces on
a particle is equivalent to the effective force
of the particle.
• The system of external and internal forces
acting on the entire system of particles is
equivalent to the system of effective forces.
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Vector Mechanics for Engineers: Dynamics
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Application of Newton’s Laws. Effective Forces
14 - 6
• Summing over all the elements,
n
iiii
n
i
n
jiji
n
iii
n
iii
n
i
n
jij
n
ii
amrfrFr
amfF
11 11
11 11
• Since the internal forces occur in equal
and opposite collinear pairs, the resultant
force and couple due to the internal
forces are zero,
iiiii
iii
amrFr
amF
• The system of external forces and the
system of effective forces are
equipollent but not equivalent.
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Linear & Angular Momentum
14 - 7
• Linear momentum of the system of
particles,
n
iii
n
iii
n
iii
amvmL
vmL
11
1
• Angular momentum about fixed point O
of system of particles,
n
iiii
n
iiii
n
iiiiO
n
iiiiO
amr
vmrvmrH
vmrH
1
11
1
• Resultant of the external forces is
equal to rate of change of linear
momentum of the system of
particles,
LF
OO HM
• Moment resultant about fixed point O of
the external forces is equal to the rate of
change of angular momentum of the
system of particles,
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Motion of the Mass Center of a System of Particles
14 - 8
• Mass center G of system of particles is defined
by position vector which satisfiesGr
n
iiiG rmrm
1
• Differentiating twice,
FLam
Lvmvm
rmrm
G
n
iiiG
n
iiiG
1
1
• The mass center moves as if the entire mass and
all of the external forces were concentrated at
that point.
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Angular Momentum About the Mass Center
14 - 9
G
n
iii
n
iiii
G
n
ii
n
iiii
n
iGiii
n
iiiiG
n
iiiiG
M
Framr
armamr
aamramrH
vmrH
11
11
11
1
• The angular momentum of the system of
particles about the mass center,
• The moment resultant about G of the external
forces is equal to the rate of change of angular
momentum about G of the system of particles.
• The centroidal frame is not,
in general, a Newtonian
frame.
• Consider the centroidal frame
of reference Gx’y’z’, which
translates with respect to the
Newtonian frame Oxyz.
iGi aaa
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Angular Momentum About the Mass Center
14 - 10
• Angular momentum about G of particles in
their absolute motion relative to the
Newtonian Oxyz frame of reference.
GGG
n
iiiiG
n
iii
n
iiGii
n
iiiiG
MHH
vmrvrm
vvmr
vmrH
11
1
1
• Angular momentum about G of
the particles in their motion
relative to the centroidal Gx’y’z’frame of reference,
n
iiiiG vmrH
1
GGi vvv
• Angular momentum about G of the particle
momenta can be calculated with respect to
either the Newtonian or centroidal frames of
reference.
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Conservation of Momentum
14 - 11
• If no external forces act on the
particles of a system, then the linear
momentum and angular momentum
about the fixed point O are
conserved.
constant constant
00
O
OO
HL
MHFL
• In some applications, such as
problems involving central forces,
constant constant
00
O
OO
HL
MHFL
• Concept of conservation of momentum
also applies to the analysis of the mass
center motion,
constant constant
constant
00
GG
G
GG
Hv
vmL
MHFL
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Concept Question
14 - 12
Three small identical spheres A, B, and C,
which can slide on a horizontal, frictionless
surface, are attached to three 200-mm-long
strings, which are tied to a ring G. Initially, each
of the spheres rotate clockwise about the ring
with a relative velocity of vrel.
Which of the following is true?
a) The linear momentum of the system is in the positive x direction
b) The angular momentum of the system is in the positive y direction
c) The angular momentum of the system about G is zero
d) The linear momentum of the system is zero
vrel
vrel
vrel
x
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Sample Problem 14.2
14 - 13
A 10-kg projectile is moving with a
velocity of 30 m/s when it explodes into
2.5 and 7.5-kg fragments. Immediately
after the explosion, the fragments travel
in the directions qA = 45o and qB = 30o.
Determine the velocity of each fragment.
SOLUTION:
• Since there are no external forces, the
linear momentum of the system is
conserved.
• Write separate component equations
for the conservation of linear
momentum.
• Solve the equations simultaneously
for the fragment velocities.
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Sample Problem 14.2
14 - 14
SOLUTION:
• Since there are no external forces, the
linear momentum of the system is
conserved.
x
y
• Write separate component equations for
the conservation of linear momentum.
mA vA +mB vB =mv0
2.5( )vA + 7.5( )vB = 10( )v0
x components:
2.5vA cos45°+ 7.5vB cos30° =10 30( )
y components:
2.5vA sin45°- 7.5vB sin30° = 0
• Solve the equations simultaneously for the
fragment velocities.
62.2m s 45 29.3m s 30A Bv v= =a c
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Group Problem Solving
14 - 15
In a game of pool, ball A is moving with a
velocity v0 when it strikes balls B and C,
which are at rest and aligned as shown.
Knowing that after the collision the three
balls move in the directions indicated and
that v0 = 4 m/s and vC= 2 m/s, determine
the magnitude of the velocity of (a) ball A,
(b) ball B.
SOLUTION:
• Since there are no external forces, the
linear momentum of the system is
conserved.
• Write separate component equations
for the conservation of linear
momentum.
• Solve the equations simultaneously
for the pool ball velocities.
vCvA
v0 vB
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Group Problem Solving
14 - 16
Write separate component equations for the conservation of
linear momentum
(4)cos 30 sin 7.4 sin 49.3 (2)cos 45
0.12880 0.75813 2.0499
A B
A B
m mv mv m
v v
m(4)sin 30° =mvA
cos 7.4°-mvB
cos 49.3° +m(2)sin 45°
0.99167vA
-0.65210vB
= 0.5858
0.12880 0.75813 2.0499A Bv v
0.99167vA
-0.65210vB
= 0.5858
x:
y:
Two equations, two unknowns - solve
0.65210 (
+ 0.75813 (
)
)
0.83581 1.7809Av
2.13 m/sAv
(1)
(2)
Sub into (1) or (2) to get vB2.34 m/sBv
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Concept Question
14 - 17
In a game of pool, ball A is moving with a
velocity v0 when it strikes balls B and C,
which are at rest and aligned as shown.
vCvA
v0 vB
After the impact, what is true
about the overall center of mass
of the system of three balls?
a) The overall system CG will move in the same direction as v0
b) The overall system CG will stay at a single, constant point
c) There is not enough information to determine the CG location
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Kinetic Energy
14 - 18
• Kinetic energy of a system of particles,
n
iii
n
iiii vmvvmT
1
221
121
iGi vvv
• Expressing the velocity in terms of the
centroidal reference frame,
n
iiiG
n
iii
n
iiiGG
n
ii
n
iiGiGi
vmvm
vmvmvvm
vvvvmT
1
2
212
21
1
2
21
1
2
121
121
• Kinetic energy is equal to kinetic energy of
mass center plus kinetic energy relative to
the centroidal frame.
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Work-Energy Principle. Conservation of Energy
14 - 19
• Principle of work and energy can be applied to each particle Pi ,
2211 TUT
where represents the work done by the internal forces
and the resultant external force acting on Pi .ijf
iF21U
• Principle of work and energy can be applied to the entire system by
adding the kinetic energies of all particles and considering the work
done by all external and internal forces.
• Although are equal and opposite, the work of these
forces will not, in general, cancel out.jiij ff
and
• If the forces acting on the particles are conservative, the work is
equal to the change in potential energy and
2211 VTVT
which expresses the principle of conservation of energy for the
system of particles.
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Principle of Impulse and Momentum
14 - 20
21
12
2
1
2
1
LdtFL
LLdtF
LF
t
t
t
t
21
12
2
1
2
1
HdtMH
HHdtM
HM
t
tO
t
tO
OO
• The momenta of the particles at time t1 and the impulse of the forces
from t1 to t2 form a system of vectors equipollent to the system of
momenta of the particles at time t2 .
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Sample Problem 14.4
14 - 21
Ball B, of mass mB,is suspended from a
cord, of length l, attached to cart A, of
mass mA, which can roll freely on a
frictionless horizontal tract. While the
cart is at rest, the ball is given an initial
velocity
Determine (a) the velocity of B as it
reaches it maximum elevation, and (b)
the maximum vertical distance h
through which B will rise.
.20 glv
SOLUTION:
• With no external horizontal forces, it
follows from the impulse-momentum
principle that the horizontal component
of momentum is conserved. This
relation can be solved for the velocity of
B at its maximum elevation.
• The conservation of energy principle
can be applied to relate the initial
kinetic energy to the maximum potential
energy. The maximum vertical distance
is determined from this relation.
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Sample Problem 14.4
14 - 22
SOLUTION:
• With no external horizontal forces, it follows from the
impulse-momentum principle that the horizontal
component of momentum is conserved. This relation can
be solved for the velocity of B at its maximum elevation.
21
2
1
LdtFLt
t
(velocity of B relative to
A is zero at position 2)2,2,2,2,
01,1, 0
AABAB
BA
vvvv
vvv
Velocities at positions 1 and 2 are
2,0 ABAB vmmvm
02,2, vmm
mvv
BA
BBA
x component equation:
2,2,1,1, BBAABBAA vmvmvmvm x
y
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Sample Problem 14.4
14 - 23
• The conservation of energy principle can be applied to relate
the initial kinetic energy to the maximum potential energy.
2211 VTVT
Position 1 - Potential Energy:
Kinetic Energy:
Position 2 - Potential Energy:
Kinetic Energy:
glmV A1
202
11 vmT B
ghmglmV BA 2
22,2
12 ABA vmmT
ghmglmvmmglmvm BAABAAB 22,2
1202
1
g
v
mm
mh
BA
A
2
20
2
0
20
22,
20
2222
v
mm
m
mg
mm
g
v
g
v
m
mm
g
vh
BA
B
B
BAA
B
BA
g
v
mm
m
g
vh
BA
B
22
20
20
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Sample Problem 14.5
14 - 24
Ball A has initial velocity v0 = 3 m/s
parallel to the axis of the table. It hits
ball B and then ball C which are both at
rest. Balls A and C hit the sides of the
table squarely at A’ and C’ and ball B
hits obliquely at B’.
Assuming perfectly elastic collisions,
determine velocities vA, vB, and vC with
which the balls hit the sides of the table.
SOLUTION:
• There are four unknowns: vA, vB,x, vB,y,
and vC.
• Write the conservation equations in
terms of the unknown velocities and
solve simultaneously.
• Solution requires four equations:
conservation principles for linear
momentum (two component equations),
angular momentum, and energy.
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Sample Problem 14.5
14 - 25
x
y
ivv
jvivv
jvv
CC
yBxBB
AA
,,
SOLUTION:
• There are four unknowns: vA,
vB,x, vB,y, and vC.
• The conservation of momentum and energy equations,
yBACxB mvmvmvmvmv
LdtFL
,,0
21
0
2
212
,2
,212
212
021
2211
CyBxBA mvvvmmvmv
VTVT
HO ,1 + MO dtòå =HO ,2
- 0.6m( )mv0 = 2.4m( )mvA - 2.1m( )mvB ,y - 0.9m( )mvC
Solving the first three equations in terms of vC,
, ,3 6 3A B y C B x Cv v v v v= = - = -
Substituting into the energy equation,
( ) ( )2 2 2
2
2 3 6 3 9
20 78 72 0
C C C
CC
v v v
v v
- + - + =
- + =
vA =1.2m s vC = 2.4m s
vB = 0.6 i -1.2 j( )m s vB =1.342m s
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Group Problem Solving
14 - 26
Three small identical spheres A, B, and C, which can slide on a horizontal,
frictionless surface, are attached to three 200-mm-long strings, which are tied
to a ring G. Initially, the spheres rotate clockwise about the ring with
a relative velocity of 0.8 m/s and the ring moves along the x-axis with a
velocity v0= (0.4 m/s)i. Suddenly, the ring breaks and the three spheres move
freely in the xy plane with A and B following paths parallel to the y-axis at a
distance a= 346 mm from each other and C following a path parallel to the x
axis. Determine (a) the velocity of each sphere, (b) the distance d.
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Group Problem Solving
14 - 27
Given: vArel= vBrel = vCrel = 0.8
m/s, v0= (0.4 m/s)i , L= 200
mm, a= 346 mm
Find: vA, vB, vC (after ring
breaks), d
SOLUTION:
• There are four unknowns: vA, vB, vB, d.
• Write the conservation equations in
terms of the unknown velocities and
solve simultaneously.
• Solution requires four equations:
conservation principles for linear
momentum (two component equations),
angular momentum, and energy.
Apply the conservation of
linear momentum equation
– find L0 before ring breaks
0 (3m) 3 (0.4 ) m(1.2 m/s)m L v i i
f A B Cmv mv mv L j j i
What is Lf (after ring breaks)?
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Group Problem Solving
14 - 28
Set L0= Lf
(1.2 m/s) ( )C A Bm mv m v v i i j
A Bv v
1.200 m/s 1.200 m/sC Cv v
From the y components,
From the x components,
Apply the conservation of angular momentum equation
0( ) 3 3 (0.2m)(0.8 m/s) 0.480G relH mlv m m H0:
Hf: ( ) ( )G f A A B A CH mv x mv x a mv d
xA
Since vA= vB, and
vC = 1.2 m/s, then:
0.480 0.346 A Cm mv mv d
0.480 0.346 1.200
0.400 0.28833
A
A
v d
d v
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Group Problem Solving
14 - 29
Need another equation-
try work-energy, where
T0 = Tf
xA
Substitute in known values:
T0:
Tf:
2 20
2 2 2 20
1 1(3m) 3
2 2
3 3m [(0.4) (0.8) ] 1.200
2 2
rel
rel
T v mv
v v m m
2 2 21 1 1
2 2 2f A B CT mv mv mv
2 2 2
2
1(1.200) 1.200
2
0.480
A A
A
v v
v
0.69282 m/sA Bv v
0.400 0.28833(0.69282) 0.20024 md
Solve for d:
0.693 m/sA v
0.693 m/sB v
1.200 m/sC v
;
;
0.200 md
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Variable Systems of Particles
14 - 30
• Kinetics principles established so far were derived for
constant systems of particles, i.e., systems which
neither gain nor lose particles.
• A large number of engineering applications require the
consideration of variable systems of particles, e.g.,
hydraulic turbine, rocket engine, etc.
• For analyses, consider auxiliary systems which consist
of the particles instantaneously within the system plus
the particles that enter or leave the system during a
short time interval. The auxiliary systems, thus
defined, are constant systems of particles.
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Steady Stream of Particles. Applications
14 - 31
• Jet Engine
• Helicopter
• Fluid Stream Diverted by Vane
or Duct
• Fan
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Steady Stream of Particles
14 - 32
• System consists of a steady stream of particles
against a vane or through a duct.
BiiAii
t
t
vmvmtFvmvm
LdtFL
21
2
1
• The auxiliary system is a constant system of
particles over t.
• Define auxiliary system which includes
particles which flow in and out over t.
AB vvdt
dmF
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Streams Gaining or Losing Mass
14 - 33
• Define auxiliary system to include particles
of mass m within system at time t plus the
particles of mass m which enter the system
over time interval t.
21
2
1
LdtFLt
t
• The auxiliary system is a constant system of
particles.
udt
dmFam
udt
dm
dt
vdmF
vmvvmvmtF
vvmmtFvmvm
a
a
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Sample Problem 14.6
14 - 34
Grain falls onto a chute at the rate of
120 kg/s. It hits the chute with a
velocity of 10 m/s and leaves with a
velocity of 7.5 m/s. The combined
weight of the chute and the grain it
carries is 3000 N with the center of
gravity at G.
Determine the reactions at C and B.
SOLUTION:
• Define a system consisting of the mass
of grain on the chute plus the mass that
is added and removed during the time
interval t.
• Apply the principles of conservation of
linear and angular momentum for three
equations for the three unknown
reactions.
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Sample Problem 14.6
14 - 35
SOLUTION:
• Define a system consisting of the
mass of grain on the chute plus the
mass that is added and removed
during the time interval t.
• Apply the principles of conservation
of linear and angular momentum for
three equations for the three
unknown reactions.
10sin
10cos
21
ByA
Bx
vmtBWCvm
vmtC
LdtFL
HC ,1 + MC dtòå = HC ,2
-1.5 Dm( )vA + -3.5W + 6B( )Dt
= 3 Dm( )vB cos10°- 6 Dm( )vB sin10°
Solve for Cx, Cy, and B with
120 kg/sm
t
D=
D
B = 2340 N C = 886 i +1704 j( ) N
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Group Problem Solving
14 - 36
The helicopter shown can produce a maximum downward air speed
of 25 m/s in a 10-m-diameter slipstream. Knowing that the weight of
the helicopter and its crew is 18 kN and assuming r= 1.21 kg/m3 for
air, determine the maximum load that the helicopter can lift while
hovering in midair.
SOLUTION:
• Calculate the time rate of change of the
mass of the air.
• Determine the thrust generated by the
airstream.
• Use this thrust to determine the
maximum load that the helicopter can
carry.
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Group Problem Solving
14 - 37
SOLUTION:
Given: vB = 25 m/s, W= 18000 N, r= 1.21 kg/m3
Find: Max load during hover
Calculate the time rate of change (dm/dt) of the mass of the air.
Choose the relationship you will
use to determine the thrust
( )B A
dmF v v
dt
mass density volume density area length
( )Bm A lr ( )B BA v tr
D
DB B
m dmA v
t dt r AB is the area of the slipstream
vB is the velocity in the slipstream.
Well above the blade, vA ≈ 0
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Group Problem Solving
14 - 38
Use statics to determine the maximum payload during hover
Use the relationship for dm/dt
to determine the thrust
( )B A
dmF v v
dt B B
dmA v
dt g
0y H PF F W W
F
WH WP
– 59,396 – 8000 41396 NP HW F W W = 41400 N
F =dm
dt( vB – vA)
= (1.21 kg/m3)p
4( )(10 m)2 ( 25 m/s)2
= 59,396 N
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Vector Mechanics for Engineers: Dynamics
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Concept Question
14 - 39
In the previous problem with the
maximum payload attached,
what happens if the helicopter
tilts (or pitches) forward?
a) The area of displaced air becomes smaller
b) The volume of displaced air becomes smaller
c) The helicopter will accelerate upward
d) The helicopter will accelerate forward
*The helicopter will also accelerate downward